Thursday, October 05, 2017

An evergreen of denial is that a colder object can never make a warmer object hotter. That's the Second Law of Thermodynamics, so according to the Agendaists, the Greenhouse Effect, with greenhouse gases playing the role of the colder object, is rubbish. They neglect the fact that heating and cooling are dynamic processes and thermodynamics is not.

Eli, of course, is a dynamic bunny and knows how to add and subtract. Divide is also possible. What is happening is that one does not have just a hot body and a cold body, but a really hot body, the sun, constantly heating a colder (much), but still warm body the Earth, which then radiates the same amount of energy to space.

Today on twitter, Eli stepped through the simple math and he thought it would be a good thing to put the thread on this blog for future reference. We start with a simple case, imagine the Earth is just a plate in space with sunlight shining on it. Maybe 400 W/m^2

The sun warms the plate, but as the plate warms it radiates until the radiated heat matches the heat being absorbed from the sun

Using the Stefan Boltzman Law you can calculate the temperature of the plate when it reaches equilibrium (400 W/m2) = 2 σ Teq4 where σ is the Stefan Boltzmann constant 5.67 x 10-8 W/(m2 K4), factor of 2 for a two sided plate per m2. Run the numbers Teq=244 K.

Now lets add another plate. We'll color this plate green for greenhouse. It is heated by the first at a rate of 200 W/m2

But after a while, it too has to heat up and reach an equilibrium temperature. . . so as a first guess something like

That's wrong though because there are 400 W/m^2 going into the two plate system and 300 coming out. At equilibrium an equal amount of energy has to be going in as coming out So what happens??

The entire system has to heat up to reach the equilibrium condition. T1 and T2 are the equilibrium temps of the plates.

Looking at the two plate system, the energy going in is 400 W/m2 and the energy going out is σT14 + σT24 Since these will be equal at equilibrium

400 W/m2 = σ T14 + σ T24

And there also has to be an equilibrium for the energy going in and out of the green plate

σ T14 = 2 σ T24

The bunnies can rearrange the second equation to get

σ T24 = 1/2 σ T14

and substitute for σ T24 back into the first equation

400 W/m2 = σ T14 + 1/2 σ T14

or

400 W/m2 = 3/2 σ T14

Solving for T1 the answer is T1 = 262 K.

Without the greenhouse plate it was 244 K.

Introduction of the second plate raised the equilibrium temperature of the first by 18 K.

The Green Plate Effect

Show this to the next fool with an agenda who thinks that the Green Plate Effect violates the Second Law of Thermodynamics

Talking about plates in space, a colleague of mine (who sadly died before this was announced) was given a NASA award for his work on passive cooling of infrared space telescopes. This basically uses sun shields and radiators to cool a space telescope without needing cryogenic coolants.

Ah, another version of https://en.wikipedia.org/wiki/Idealized_greenhouse_model or http://scienceblogs.com/stoat/2014/02/16/the-idealised-greenhouse-effect-model-and-its-enemies/ but with nicer pictures. Surely *this time* people will understand :-)

My entry on Spencers blog I think proves with a physical object that coll objects can make a warm object warmer:

September 25, 2017 at 11:10 AM

Using whatever version of physics you believe in please explain how a room temperature thermal imaging camera can take thermographs of objects down to -50C There are no “cold rays” so how does focusing an object at -40C onto a microbolometer at 25C change its temperature compared to focussing an object at -30C onto microbolometer at 25C?https://3.bp.blogspot.com/-GT_Ar-9WWfQ/UNkU2Fb2nBI/AAAAAAAAA1M/NLxj8Rt7yRI/s1600/sky+high+low+cloud.jpg This shows a thermal image of sky and clouds (in winter). The camera body was approx. 15 to 20C and the sensor was uncooled and therefore at a similar temperature perhaps hotter. If you are correct and cold objects cannot cause warmer objects to warm then this picture should effectively be an even black (there is no object above the microbolometer temperature and in your view lower temperatures cannot warm hotter objects and remember there are no such things as cold rays). All objects in the field are less warm than the sensor and in your physics cannot change the temperature of the sensor. However you can see temperatures from -2C to -35C. This is because the cold clouds are providing energy to the sensor changing its temperature. If the object in the field of view were at abs zero then no heat would be transferred to the sensor and the sensor would be in equilibrium with its camera environment. Above absolute zero heat energy is transferred to the sensor and its temperature increases above the camera background. At -50C (approx.) the heat from the object adds sufficient energy to the sensor for detection and the sensor warms until the energy OUT to the camera environment equals the energy IN from the camera environment plus the energy in through the lens. This additional energy changes in microbolometer temperature and hence its resistance. You should also be aware that the germanium lens used on thermal imaging cameras acts as a bandpass filter. So most radiation from CO2 and water vapour is not passed to the sensor. Hence no back radiation is seen from water vapour or CO2. Clouds of course are not water vapour but a cloud is an aerosol comprising a visible mass of minute liquid droplets (wiki) and hence thermal radiation is more like a black body allowing the camera to see the cloud. In general a thermal imaging camera MUST be insensitive to GHG radiation otherwise hot air would fog the image.

Wm, the point is to have something in big print and simple figures that can be pointed to in a tweet. Understanding from those with an agenda is not hoped for. Stopping them from confusing the crowd might be

It may interest andthentheresphysics to know that the same emissivity games could work to the advantage of ground based IR telescopes - you can tone down the thermal infrared background noise by slathering your instrument and the insdide of the dome with paint spiked with boron nitride, which though isoelectronic with graphite is a wide bandgap semiconductor whose strong bonds and light atoms make it an almost perfect IR reflector in the 10-13 micron range, which tears a hole in its emissivity in that range

In the same way, I suppose, one does not add up 400 w/m2 from each side of the sun to arrive at 800 w/m2.

With my limited abilities, I suspect I'm about to be told that the spherical sun does indeed emit more than 400 w/m2 (as received by the inside surface of a sphere surrounding the sun), and that this is the amount received from the surface of the sun facing the plate.

Eli set up a non rotating plate with the sun on one side so the sun only shines on one side. That's the model. The condition was chosen to simplify the discussion and avoid the need to discuss the effect of rotation of the Earth and the fact that the Earth is spherical

Should a bunny look at a more complex discussion such as the Weasel pointed to in Wikipedia or Arthur Smith published on arXiv he or she would find that the solar Flux is ~1400 W/m^2 on a surface perpendicular to the distance btw Earth and Sun. Because the sun shines only on one side at a time and the fact that the Earth is a sphere the average intensity integrated over the sphere is 1/4 of that or about 350 W/m^2.

This may be an uninformed question, but has an analogous physical experiment been done to demonstrate this. It seems that one should easily be able to set up a radiative source with 2 plates and measure the temperatures. Would that be even more convincing for the 'skeptics'?

Ken, this was well known even before 1900 so there is little publlshed work as such. However, it is inherent in the design of vacuum furnaces and cryostats. Cryostats are the inverse. You have a cooling unit with heat shields around it that keep the heat from the outside leaking in. Eli hisself operated a vacuum oven as a grad student with a several heat shield around the heating element and if they were not there the temperature of the oven was much lower than what was needed.

Eli, you've mistaken me for my opponent at the blog I linked to. A contrarian there replied that the blue plate should emit 400 w/m2 on both sides. Knowing little of physics I overthought what he said and wasn't confident he was wrong, so came here to ask for help. The 2-bulb thing was what I came up with on my own to uncomplicate my thinking and see his error. Seems I haven't expressed myself clearly.

I've saved this page for future reference. It's the tidiest rebuttal to the simplistic 2nd Law mantra I've ever seen.

So the Bunny went over and looked at the food fight about this post that has broken out at Roy's place, and indeed Barry is right, he is doing good work.

Eli has taken down a couple of the comments above which he was mistaken in attributing the ungodly position to Barry, but repeats them here for content

Most importantly the post was to show the placing a colder body near a warmer body can make the warmer body hotter when it is being heated by another source.

The relevance of this proof is that the Earth is heated by light from the Sun and layers of the atmosphere containing greenhouse gasses act as the colder third body.

What has been proved is that colder bodies can warm a hotter body through back radiation. Where there is an external or internal heat source in the system, the equilibrium temperature of the warm body will be higher than if the cold body were not present.

Yes as Gerlach & Tsheuscner point out the solar Flux at the Sun is much higher than the IR Flux from the Earth at the Earth but the solar flux at the Earth's surface matches the Earth's IR flux because of things like the 1/r2 falloff with distance from the Sun, the fact that only a half of the Earth is illuminated at a time by the Sun and the fact that the Earth is roughly spherical so that most of the surface is tilted to the impinging light from the Sun.

But anybunny claiming that the Sun has to heat both sides of the blue plate at the same time is claiming a binary star illuminating both sides of the Earth. There is this thing called night.

As to light plates, well Eli might ask folks to look at their cell phones. Most of them have flashlight apps.

"This principle on which the whole following development is based is : heat can never pass from a cooler to a warmer body, unless another change associated herewith simultaneously happens."

Moreover as Bindidon points out Clausius was well aware of what we call back radiation:

"What further regards heat radiation as happening in the usual manner, it is known that not only the warm body radiates heat to the cold one but that the cold body radiates to the warm one as well, however the total result of this simultaneous double heat exchange is, as can be viewed as evidence based on experience, that the cold body always experiences an increase in heat at the expense of the warmer one."

Both conditions are met in the Green Plate effect.

You appear to have missed the point that in this proof the sun (or at least another body) is shining on the blue plate transferring heat to it continuously.

If you just have a warm body sitting in space, it will cool by radiation.

If you have a warm body and a colder one near it, it will cool a little bit slower because of interchange of radiative energy btw the two The net interchange of heat will be from the warmer to the cooler. But in both cases the body(s) will cool down to the background temperature of the universe, like a few C

However, and here is what folk miss, if you have a heat source, like the sun, heating the warm body at a constant rate while it cools by radiation, the warm body will become hotter if there is a colder body near it because of the interchange of radiative energy between them.

Joseph Postma claims that because the blue plate example always emits less than 400W back at the energy source (no matter how many green plates are added), this example refutes the RGHE. Of course this is idiotic because the blue plate does emit more than 400W if you consider both sides of it.

As thefordprefect notes above, how does a standard, easily purchased thermal imaging camera (e.g., http://www.flirmedia.com/MMC/THG/Brochures/IND_025/IND_025_EN.pdf) take pictures down to -40C with a room temperature sensor without violating your imaginary laws?

How does it take a picture of anything at all cooler than the sensor, say even at 18C?

I hope that it is OK to present an alternate interpretation to points made civilly.

I think Betty is right !

A temperature sensing device has a flux incident upon it.

It has an electrical current and a programmed "computer" that converts the incident flux into a temperature value for display.

Surely any value can be sensed ? As Betty said the flow of heat is from the sensor to the colder object !

It does not have to induce warming - it is the calculations using radiation physics that determines the temperature.

In no way does this involve the transfer of heat from a cold object !

Hence thefordprefect's argument is incorrect.

Someone once claimed a microwave oven proved that a cold object could heat a warmer object but that is just silly.

Google Pictet's reflection of cold to prove for yourself that the radiation from a cold object does not increase the temperature of a warmer object !

Similarly there are thousands of real world examples where the input of continuous energy does not necessarily involve increases in temperature.

The best know is water - at STP no matter how much energy you "throw" at water it will never increase in temperature above 100°C.

At even lower temperatures such as ocean surfaces water evaporates without the mass of water necessarily increasing in temperature.

The hottest land surface temperature recorded is 70.7°C whilst the natural hottest ocean temperature is ~36°C.

Similarly for melting of all objects that melt.

Viscosity of liquids decreases as they are "heated" and the temperature changes are NOT in accordance with the specific heats - I did experiments such as these in Chemistry classes at University 46 years ago.

The Stefan-Boltzmann, Planck's, and Wein's laws arose from the cavity radiation experiments and involved the emission of continuous spectra. No-one knows for certain that these laws apply to circumstances other than the emission of continuous spectra.

No one ever said the green plate was warming the blue plate. The sun is doing the warming. Greenhouse deniers consistently miss this very simple point. All the green plate is doing is affecting the radiative cooling rate of the blue plate as Eli has been at pains to point out.

Ross McLeod writes: "I hope you can accept alternate viewpoints to open discussion."

This is math and physics. There really isn't an alternate view. "Views on shape of the earth differs" doesn't work here.

Eli has provided a very simple system diagram complete with the necessary numbers. It's either correct or it's incorrect. In your comments you never once show where it is incorrect. If you could, I'm sure you would have.

The question then becomes: what do *you* do in the face of evidence that proves your position incorrect? You can *deny* that evidence or you can accept it and modify your view accordingly.

I found some time to create an Excel Spreadsheet using the EXACT algebra used in the post.

Planck's equation is the only equation that completely describes any blackbody radiant emission and the corresponding temperature relationship. The Stefan-Boltzmann equation describes the total power emitted at any temperature.

We can all agree on this surely ?

The area under the curve plotted using Planck's equation multiplied by pi is equal to the Stefan-Boltzmann calculated total power for the corresponding temperature.

You do realize that you are specifically saying that whether you put a black body plate in front of your face or not, you still feel the exact same radiation from the sun, right? In other words, you are stating that said black body plate acts perfectly transparent to radiation.

Are you entirely sure this is what you really want to say? Black bodies were never assumed to be invisible to all incident radiation in my books.

The only way for this to happen is for plate 1 to be transparent or effectively so by emitting the same 400W/m^2 it receives on the exposed side on the nonexposed side as well. That is you are saying that for plate 2 the above situation and:

SUN -->> transparent plate -->> plate 2 (temp X)

are completely equivalent. They are not. This is because plate 1 does NOT emit 400 W/m^2 from both sides equally.

If 400 W/m2 come into the system then 400 W/m2 go out from the ENTIRE system which is what this example. To avoid a valid accusation of politically convenient denial of conservation if energy please tell Eli how much thermal energy is coming out of the sun side of the blue plate, the green plate facing side of the blue plate, the blue plate facing side of the green plate and the space facing side of the green plate.

The answer may surprise you but it is the same process which explains how the surface of Venus is so hot

This is provable mathematically using Planck's equation and the relationship between the Stefan-Boltzmann, Planck's and Wein's law!

The ONLY empirical evidence that exists as a result of the cavity oven experiments in the 19th century is that a black body emits radiation in proportion to its temperature.

No empirical evidence exists for any other algebraic manipulations of the Stefan-Boltzmann law !

If such algebraic manipulations of the type performed in this post were VALID then they would be verifiable using Planck's law AND THEY DO NOT SATISFY this necessary requirement.

Simple algebraic sums involving the Stefan-Boltzmann law are not valid so all of the huffing and puffing combined with insulting one another is totally pointless.

I find Betty's points totally supportable !

The standard argument for the greenhouse effect is that the "atmospheric layer" emits the same radiation on both sides in accordance with its temperature because it is a layer.

This post turns that argument on its head - can't you see that ?

Besides it is not valid to sum radiative flux using the Stefan-Boltzmann equation and calculate the final temperature.

This is not saying that there is any problem with IR thermometers as they do not use these algebraic manipulations.

I'll say it again - EVERY time you see some form of algebraic manipulation of the Stefan-Boltzmann equation such as performed here check it out using Planck's law.

If for example the equality as stated in the post -

σ T(1)4 = 2 σ T(2)4 AND

T(1) = ~262 K - which means T(2) = ~220.

Then plotting a Planck curve for T(1) - 220 K - and then multiplying every value by 2 MUST produce a curve for T(2) - 262 K - THE AUTHOR HAS SAID THIS IS TRUE !!!

BUT IT DOESN'T !!

The 2 x T(1) curve does indeed have an area under the curve of 266.66 satisfying the numerical requirement of the Stefan-Boltzmann equation but IT is not a true Planck curve and therefore this algebraic manipulation cannot be correct !

This is the crucial point.

As temperature increases the wavelength of the peak emission shifts to shorter wavelengths and there is no way to account for this FACT using the Stefan-Boltzmann equation ! Further, hot objects emit radiation at shorter wavelengths where the emission from a cooler object is near or zero.

This in no way implies there is anything wrong with the Stefan-Boltzmann equation or its useful applications - it is, however, stating categorically that there is a major problem with the way people use it using simple algebra !

Have a look at either the SB or Planck's equations - why would you expect there to be simple algebraic relationships ?

Try it for yourself using a spread sheet and plot some curves - you will see this is right.

If you had just the blue plate, and the other side of it was perfectly insulated, then yes, 400 W/m2 would warm it to 290K, and it would radiate 400 W/m2 back. If you then removed the insulation, the other side would also start radiating 400 W/m2. Because the plate is now radiating 800 W/m2, and only absorbing 400 W/m2, it can't stay in equilibrium and will then start cooling. It will reach equilibrium at 244K (absorbing 400 W/m2, radiating 200 W/m2 from both sides).

With Ross McLeod, I can't even tell what position he's defending. Is it:

a) With just the blue plate, it reaches 244K. When you add the green plate, both reach a temperature of 244K

b) With just the blue plate, it reaches 290K. When you add the green plate, both reach a temperature of 290K

c) Something else?

Over at Climate of Sophistry, I think they've decided on option b. Or, at least, the 2nd half of option b; I'm not sure what they think happens with just the blue plate.

Eli Rabrtt said..."Energy is conserved. Temperature is not a conserved quantity. Besides which where did 290k come from? You need to get off those funny little pills"

YOU are conserving FLUX - W/m2 and that is most definitely NOT a conserved quantity !

If you look at the "net" form of the SB equation for the exchange of energy between 2 objects at T(1) and T(2) -

P(net) = σ (T(1)^4 - T(2)^4) -

the quantity in the brackets is heat - the energy exchanged between them due to the temperature difference.

If P(net) is positive T(1) is decreasing and presumably T(2) may increase.

If P(net) is negative then T(1) is increasing as T(2) decreases.

If they have the same temperature there is no exchange of heat and we call that thermal equilibrium.

Flux is not a conserved quantity.

"where did 290k come from?"

400/sigma = T^4 T = 290 K !

Nothing funny about that at all - the presence or absence of pills or otherwise !!

"The answer may surprise you but it is the same process which explains how the surface of Venus is so hot"

Absolute rubbish !

Venus is so hot because it has a huge number of active volcanoes continuously belching CO2, SO2, CH4 etc into it's already dense atmosphere such that it has a surface pressure of ~92 bar.

Sunlight doesn't even reach Venus's surfaces - the high altitude H2SO4 clouds reflect a significant portion of incident solar radiation and absorb a significant portion of the rest.

Venus albedo is at least 0.8 !

CO2 absorbs IR strongly at ~2.7 and ~4.3 micron and the Solar radiation has plenty of power at those wavelengths at Venus' orbit.

If nothing makes it to the ground it can't be a "greenhouse effect" can it ?

Harry Dale Huffman wrote this years ago - I've summarized it here:-

The Magellan spacecraft that took measurements of Venus' atmosphere as it descended to the surface before being crushed by the huge pressure ! It's readings show that the temperature high in Venus' atmosphere where the pressure is equivalent to 1 atmosphere on Earth - ~1000 millibar - is roughly ~339 K compared to ~288 K for Earth at the same pressure.

Venus is closer to the Sun - obviously. The average orbital radius of Venus is 108.21 million kilometres and for Earth it is 150 million kilometres - NASA Fact Sheets.

Using the inverse square relationships Venus receives roughly (150/108.21) squared = ~1.92 times the power per unit area that Earth receives.

Using the SB law Earth's ~288 K at one atmosphere - 1000 millibar - is raised by the 4th root of ~1.92 = 1.177 times ~288 = ~339 K which is the temperature the Magellan spacecraft recorded at equivalent pressures !

The problem with this is that the area radiating from the blue plate is twice as large as the area being illuminated. Only one side of the blue plate is being irradiated but when both sides are at some temperature T both sides will radiate energy. This would be the solution to a problem where the blue plate had only a single side.

Thus the proper equation

400 W/m2 x A = 2 A Sigma T^4

The area of the blue plate on both sides of the equation cancel.

If you want to change the problem, you have a different problem

Everything that followed from the wrong statement of the problem is chaff.

Ross says: "There is modern experimental evidence which verifies the 2 centuries old experiment demonstrating that the radiation from a cold object does not induce warming in a warmer object !"

Please desist with mis-stating what was discussed in the post (Do it again and you are a simple liar). What was shown is that when a surface is heated at a constant rate, introduction of a second surface which captures some of the thermal radiation from the first surface and re-radiates a portion to the first will raise the equilibrium temperature of the heated surface.

In spite of your thrashing around, this does not violate the second law of thermodynamics and can be used to illuminate the role of greenhouse gases in maintaining the Earth's surface temperature.

The sensor detects thermal energy flowing from it to the colder object."

On NET as Clausius said:

"What further regards heat radiation as happening in the usual manner, it is known that not only the warm body radiates heat to the cold one but that the cold body radiates to the warm one as well, however the total result of this simultaneous double heat exchange is, as can be viewed as evidence based on experience, that the cold body always experiences an increase in heat at the expense of the warmer one."

'There is another version of the second law and I think the scenario violates this one as well.

No process is possible whose sole result is the transfer of heat from a cooler to a hotter body."-----------------------------------------------------You left out the fact that one side of the blue plate is being heated by a power source. And as Clausius said (see above)

"What further regards heat radiation as happening in the usual manner, it is known that not only the warm body radiates heat to the cold one but that the cold body radiates to the warm one as well, however the total result of this simultaneous double heat exchange is, as can be viewed as evidence based on experience, that the cold body always experiences an increase in heat at the expense of the warmer one."

I say I have a spreadsheet which proves that all of the algebraic manipulations on this post are incorrect. None of them produce valid Planck curves for the sums etc of differing radiation values - NONE of them !

I'll supply this to anyone who wants it - it is ~2.2 MB and opens in Excel or OpenOffice or LibreOffice. (I guess it is up to the moderators to allow an exchange of documents etc.)

"Ross says: "There is modern experimental evidence which verifies the 2 centuries old experiment demonstrating that the radiation from a cold object does not induce warming in a warmer object !"

Please desist with mis-stating what was discussed in the post (Do it again and you are a simple liar)."

I cannot believe you can be so angry that you call someone a liar for stating a simple TRUE FACT !

I said - "There is modern experimental evidence which verifies the 2 centuries old experiment demonstrating that the radiation from a cold object does not induce warming in a warmer object !"

This is FACT and it is TRUE. Pictet proved this more than 200 years ago !

You ARE claiming it is the back radiation from the green plate that makes the blue plate hotter no matter how you obfuscate your claims so this is relevant !!!

I simply made a factual statement about a 200 year old experiment that was replicated in February 1984 and the historical context of the science of the late 1790's early 1800's that lead to the original experiment was published after being accepted for re-publication in the American Journal of Physics in August 1985 !

I still say that experiment trumps thought bubble any day !

This document is available here :-

http://www2.ups.edu/physics/faculty/evans/Pictet%27s%20experiment.pdf

I am decidedly and provably NOT a liar and I do not "drop my bundle" when challenged !

I still say disprove my claims about the spreadsheet proving the algebra used in this post is wrong !

Betty, Eli's plates are perfectly black and absorb all radiation that falls on one side. They don't have to be one molecule thick just their area A >> t^2 so edge effects don't matter and have high thermal conductivity so the temperature of one side is equal to the temperature of the other.

Betty, the Stefan Boltzman relationship was derived empirically by Stefan and theoretically by Boltzmann from thermodynamics https://physics.stackexchange.com/questions/319861/boltzmann-s-original-derivation-of-the-stefan-boltzmann-law

A one-molecule thick wall would not behave as a black body, so no, that's not the assumption. Light would sail through it mostly unimpeded, it would be transparent. Assume plate is thick enough to behave as a black body (the thickness of a sheet of alfoil would be enough) and thin enough that any temperature difference between the two sides is negligible. Such a plate does have two sides.

"What gives you permission to add radiative fluxes?"

That would be the principle that says that blackbodies absorb all radiation falling upon them. Read the definition of a blackbody.

"Tell us about this magical perfect insulator."

Of course there's no perfect insulator. If it will make you happier, assume a good insulator (e.g. aerogel) thick enough that conductive loss can be ignored, to a first approximation. This sort of pointless nitpicking is just a waste of time.

"If the plate was an isocahedron would it be emitting 20 W/m^2 from each face?"

Why on earth would anyone think that? Obviously, if the icosahedron is at 244K, each bit of its surface will radiate at 200 W/m2, so multiply by the surface area of the icosahedron if you want the total rate of heat loss. For our thin wall, that also radiates at 200 W/m2 for the entire surface at 244 K. Since 1 m^2 of wall has 2 m^2 of surface area, it radiates 2 * 200 W/m2, in equilibrium with the incoming energy (based on the Stefan-Boltzmann equation).

How you would calculate results for such scenarios by plotting Planck curves, I can't imagine. All I'm relying upon here is the 1st law of thermo (conservation of energy), and the definition of a blackbody.

From Rosco: "Simple algebraic sums involving the Stefan-Boltzmann law are not valid".

No idea what you're talking about. Please give an example of one of these supposed sums. All I'm using S-B for is to give the power output of a surface at a given temperature.

Once again: is it possible to get a straight answer as to what you think the solution is, both for the single blue place and the blue and green plates?

Ross: Picet did NOT consider the case where one of the bodies was receiving a constant amount of thermal energy from another source as Eli described in this post. As a matter of fact, Picet's case was considered in detail by Clausius in his work on the Mechanical Theory of Heat Chapter XII, pp 335 where he points out:

"What further regards heat radiation as happening in the usual manner, it is known that not only the warm body radiates heat to the cold one but that the cold body radiates to the warm one as well, however the total result of this simultaneous double heat exchange is, as can be viewed as evidence based on experience, that the cold body always experiences an increase in heat at the expense of the warmer one."

When Pictet previously showed that a heated object at the focus of the mirror caused an increase in temperature registered by the thermometer he proved that heat radiates from hot to cold.

I note he was not the first to perform such experiments with a heated object but he was the first to perform the "reflection of cold" experiment.

When the flask of ice was situated at the focus of the second mirror the thermometer became the source of heat and it decreased in temperature.

This is entirely relevant !

This post is claiming the "back" radiation from the green plate heats the blue plate despite the green plate being at a lower temperature.

This is exactly what is being claimed !!!

I have not insulted any one and I am not making irrelevant comments. I am not "running your squirrel through Eli's patch".

I am offering realistic arguments that conflict with the claims of the post which I still say I can prove are wrong using mathematics involving Planck curves!

It doesn't matter whether or not there is a continuous amount of thermal energy input the whole claim of the post is that the back radiation from the green plate causes the blue plate to heat up !

It is deceptive to claim it isn't - the initial starting point had the input of 400 W/m2 causing the maximum effect it can on its own ! It was explicitly stated !

Quokka aid :-

"From Rosco: "Simple algebraic sums involving the Stefan-Boltzmann law are not valid".

No idea what you're talking about. Please give an example of one of these supposed sums. All I'm using S-B for is to give the power output of a surface at a given temperature."

If you can't see the inherent sums, multiples and fractions involving the SB equation in the the algebraic manipulations written as equations in the post then you must be blind !!!

The algebraic manipulations in this post CAN be shown to be mathematically incorrect using Planck curves !

There are only 2 possibilities :-

1. Algebraic sums DO NOT provide the correct answer when using the SB equation.

σT(1)^4 does not equal 2σT(2)^4.

Using the claims of the post T(1) equals the temperature of 400 W/m2 = 3/2 σT(1)^4 = ~262 K and therefore T(1) equals about 220 K.

Plot these 2 curves and then multiply every radiant emission value calculated for every increment of wavelength by 2 and plot this curve.

The curve for 2σT(2)^4 has an area under the curve equal to the area under the curve for T(1) - 266.66... W/m2 BUT it is not the same curve for the T(1) curve = 262 K !

Peak emission is not shifted to shorter wavelengths by this simple algebraic multiplication AND the curve for 2 x T(2) is NOT contained completely within the curve for T(1) as EVERY other legitimate Planck curve IS !

OR -

2. The explicit proven mathematical relationships between the Stefan-Boltzmann, Planck's and Wein's laws are invalid.

It received 400. why is it emitting 200 from each face?Oh, here you don't multiply 200 by 12? Funny that you treat it differently from a flat plate.

You are wrong.The plate you describe will still emit 400 in all directions at equilibrium (290K). And you can't add 400 to 400 to create a strawman 800 value to falsely disqualify it - otherwise you admitted an icosahedron will emit 20*200=4000 W/m2 (but receive 400 W/m2 ... cooling)

The motion of molecules at a given level of radiation. A solid plate will conduct. All the molecules will dance in tune with the incoming frequencies. The max (equilibrium) temp is 400=sigma*T^4 , T = 290K

Betty asks: "So you can add up the radiation from two bodies and get a higher temperatures? What is the temperature of two ice cubes vs one?"

Well, let's see. Ice is at 273K, so if the entire sky (as viewed from a surface) was ice, it would supply 315W. So consider a surface at 0K, facing a sky at 0K. Introduce an ice cube, occupying 1/1000th of the field of view. It will therefore supply 0.315W (on average; more at zenith, less if low on the horizon), which equates to a temperature of 48.5K (way higher than the cosmic background radiation!). A second ice cube will double the power to 0.63W, equating to 57.7K. See, not hard.

You've already invented a very special black body that can take in 400W/m^2 on one side and then emit a total of 800W/m^2 from 2 sides which could then be harvested by suitable collectors stationed above and below said special black body. I should think you'd be rich enough already. Don't be greedy.

Betty responds: "10,000 ice cubes and you get 3150W or 485K or 212C. You can boil water with 10,000 ice cubes! You just solved the world energy crisis. You must patent right away."

The flaw in this reasoning is trivially obvious, but I'll spell it out for logic-challenged. The maximum amount of energy you can get from ice, 315 W/m2, is if the entire hemispherical field of view is filled with ice (i.e. 1000 cubes). Where were you planning to put the other 9000 cubes, unless you can find a hemispherical field of view of 20*pi steradians? (cf. the usual 2*pi steradians we get in our universe)

You 'forgot' to point out any error in my calculation with 1 and 2 ice cubes. That's because you can't, of course. But please do show us your calculation of how much energy an ice cube would supply, I'm sure we'd all enjoy a laugh.

It describes how much energy is emitted from a surface, depending on its temperature. Whether that surface is on a thin plate, a thick plate, or a solid object such as a sphere is irrelevant (because they do not affect the two inputs to the SB law, temperature and surface area). But if that surface is on a plate, then we can also apply the SB law to the surface on the other side of the plate. If the whole plate is at 290K, then it tells us that both surfaces independently emit 400 W/m^2. If there is only 400 W/m^2 being absorbed, then the plate is losing 400 W/m^2 energy and therefore must cool down (due to an obscure principle, which you may even have heard of, called the Law of Conservation of Energy).

The idea that an icosahedron with 20 faces at 133C would only emit 20 W/m^2 from each face is of course ridiculous. The S-B law doesn't depend on how many planar faces there are, it cares about temperature and surface area. If we took our plate and put slight bends in it to form a concertina shape with many planar surfaces, the amount of energy it radiates would be unchanged because neither the temperature nor the surface area have changed.

The 400 W/m2 is the input to the blue plate. It could be illumination, it could be an internal electrical heater or it could be a hotter body off somewhere to the left of the diagrams. The only important thing about it is that is a constant source of heat. It is the ONLY input of energy to the system, whether the system is the blue plate alone or the blue and green plates together

The surroundings are everything EXCEPT the blue plate alone or the blue and green plates when the green plate is present

That means that at equilibrium the system has to be rejecting 400 W/m2 to the surroundings to conserve energy. The amount of energy emitted from the system per second can never exceed 400 W/m2 but it can be less while any part of the system is heating up.

Pretty much everybunny who tries to falsify the example misses or misuses this

Quokka says: "Introduce an ice cube, occupying 1/1000th of the field of view...."

Betty makes an interesting (but misguided) reply: "10,000 ice cubes and you get 3150W or 485K or 212C"

Betty, 1000 cubes that each cover 1/1000th of the field of view means the entire field of view is already covered by the first 1,000 blocks of ice. You basically have an igloo with a surface on the floor of the igloo that is receiving the radiation. The floor will receive 315 W/m^2 and will warm/cool to 0C (assuming it is insulated on the bottom side). A new ice block -- no matter where you place it, cannot cover more that 100% of the field of view and cannot provide more that 315 W/m^2 of radiation to the floor. Any blocks beyond the first 1000 will either be * on the OUTSIDE of the igloo, and provide no radiation to the floor.* on the INSIDE of the igloo, in which case they they do radiate to the floor, but simultaneously block and equal amount of radiation from the ice behind them.

This geometry seems to escape many people, and it leads people to incorrect conclusions. 10,000 blocks of ice in this situation would make the igloo's walls *thicker* but this does not provide more radiation.

Do you still claim each face emits 200? How much is each face emitting assuming the icosahedron is radiated on one face (the rest masked by "perfect" insulator). Answer the question, otherwise I will presume your old answer 20*200=4000 W/m^2.

Yes, Betty, you could and then each would only provide 1/10 as much flux as before and the whole shell would still provide the same total flux of thermal IR. And you could still not exceed 315 W/m^2 total.

It is NOT about one specific thought experiment -- it is about general principles. Whether we divide the shell in two parts or 10 parts or 1000 parts or 10000 parts, the total shell sill provides the same total power.

"We've already established that fluxes can't be added"I didn't see anyone establish this as a general principle. It is more subtle.

* Fluxes from DIFFERENT directions can and do add. If we focus two different spotlights from two different directions down onto the ground, it will be brighter than one spotlight and will warm the ground more than a single spotlight. If we focus the flux from a thousand mirrors onto a tower (like a concentrated solar thermal plant), then the fluxes add and the tower gets much hotter than the flux just of the sun.

* Fluxes from the SAME direction cannot and do not add. If one spot light was placed directly behind the first, then the spot on the ground does not get brighter (because the 2nd spotlight is merely lighting up the back of the first spotlight!).

With the ice example, the first 1000 ice cubes which each cover 1/1000 of the hemisphere can add if they are carefully placed. We can get up to 0.315 X 1000 = 315 W/m^2 of radiation onto the ground. This is the best we can do with ice. We have thermal radiation already coming from every direction. Any more such ice cubes is the equivalent of putting one spotlight behind the other.

EliRabett said..."Picot's experiment does not consider the case where the hotter body is receiving a constant input of thermal energy. Clausius analyzed it in detail. You can use Google translate.

It's squirrels all the way down with you."

Firstly it is Pictet.

Secondly your claim is nonsense.

Pictet allowed the air thermometer to come into thermal equilibrium with the air of the laboratory and that air temperature did not change to any significant extent. This was evidenced by the thermometer warming back to its initial temperature when moved away from the focus of the mirror.

Of course the thermometer is receiving a constant input of thermal from the air of the room via contact !!

And if your scenario is right when the flask of ice is placed at the second focus the thermometer is receiving more energy reflected by the two parabolic mirrors because the ice is emitting energy that was not present before it was placed there.

Thus the thermometer did indeed have a constant supply of thermal energy and placing the flask of ice at the focus supplied extra energy that was not there before.

Yet the ONLY reaction was a decrease in the temperature of the thermometer despite the extra thermal energy emitted by the ice to the thermometer.

EliRabett said..."Ross darlin, the sum of two Planck curves is not a Planck curve, just like exp(ax) + exp(bx) is not an exponential. Learn some math."

Firstly, if the only argument you have is condescension then you've damaged your position before you've even begun.

Secondly I never made any stupid arguments "like exp(ax) + exp(bx) is not an exponential"

That is a typical tactic of someone who cannot argue against a plainly stated case - misdirection to ridicule when their opponent NEVER said any such thing !

I explicitly said there is a relationship between Planck's, the Stefan-Boltzmann and Wein's equations !

The Stefan-Boltzmann equation was derived empirically BUT it is also derived from Planck's by integration - THIS IS FACT !!!

Wein's equation was derived empirically BUT it is also derived from Planck's by differentiating Planck's equation and setting the expression to zero to establish peak emissions - THIS IS FACT !!!

These FACTS tend to strengthen the validity of each equation !

Bearing these FACTS - and they are the ONLY FACTS I submit as evidence - in mind let's analyse the math quoted here.

At T(1) = ~262 K the radiant emission is 266.66... W/m2.

At T(2) the radiant emission of 133.33...W/m2 is the emission from a temperature of ~220 K.

"The bunnies can rearrange the second equation to get

σ T(2)4 = 1/2 σ T(1)4"

So plot a curve for ~220 K - or better yet program the spreadsheet to do ALL calculations for you using its maximum precision and just input the accepted values for the SB constant, Planck's and Boltzmann's constants and the SB and Planck's equations etc.

Thus the only input is 400 and the various ratios as stated in the post - 2 times, 1/2, 3/2 etc.

Thus there is no denying that at ~220 K the green plate is emitting ~133.33 W/m2 and this numeric value is indeed half of what the blue plate is emitting at ~262 K which is indeed calculated to be ~266.66 W/m2 - AND I NEVER HAVE !!!

BUT there is another consideration in the algebraic "equality" "σ T(2)4 = 1/2 σ T(1)4" that the Stefan-Boltzmann equation CANNOT illustrate - only Planck's equation can !

Plot the 2 curves - ~262 K and ~220 K.

The area under each curve is σ T(1)4 and σ T(2)4 respectively !

This is indisputable - read the reference at SpectralCalc - some people who do know math !

The laws of calculus state that the integral of 1/2f(x) is the same as 1/2 times the integral of f(x).

So if I multiply every "y" axis value (Spectral Radiance) of the T(1) curve for the corresponding "x" axis value (wavelength, wavenumber, frequency or whatever is chosen) by 1/2 then this is EXACTLY equivalent numerically to 1/2 σ T(1)4 !

This is indisputable ! If you do dispute this then you don't know math !

And the curve for this mathematical transformation should be EXACTLY equivalent to "σ T(2)4" - Eli says it is so !

But it isn't !!

And that is the point Eli darlin !!

What I say is valid mathematics and it proves your assertions about adding up flux and calculating the temperature from the algebraic sums using the SB equation are wrong !

Don't say there are no sums or subtractions in the post as that is just a ridiculous assertion primary school children can refute !

No, Ross, the air is a constant TEMPERATURE heat bath which is exposed to the thermometer. The net amount of heat flowing into the thermometer from the air depends on the temperature difference between the air and the thermometer.

Clarifying* exactly which object * exactly which surface of which object. * the orientations & positions & separations of various surfaces.

The original scenario provides clarity and is properly done. The various later ice scenarios are less clear and hence much tougher to analyze and to comment intelligently on.

As a start:* outgoing flux densities from a flat bit of surface depend only on the temperature and emissivity. It makes no sense to try to add outgoing flux densities. (for example, the outward flux density from ice @ 0 C is ~ 315 W/m* incoming flux densities to a flat bit of surface do add. You have to integrate the incoming flux densities over all the bits of solid angle.

"Quokka arbitrarily chose 1/1000th field of view. Well, I can arbitrarily choose 1/10000th, and claim he invented an igloo oven."

Yes, of course you could choose 1/10000th, either by using smaller ice cubes, or putting the ice cubes further away so that it now takes 10,000 ice cubes to cover the sky, rather than 1000. The point, which others have already made so I won't belabor it unduly, is that the maximum amount of energy that can be supplied by ice is 315 W/m^2, which happens when ice fills the entire field of view. Whether it is filled by one big ice cube close up, or trillions of them miles away, doesn't matter.

"Rather then call him out for his crackpot physics of adding fluxes, assholes actually defend him."

OK, show us your non-crackpot calculation of how much radiation an ice cube occupying 1/1000th of the sky provides. And then use that formula to calculate the radiation from two ice cubes near each other. I think you'll find that the amount of radiation doubles. Or are you really claiming that if we put a second sun in the sky, it wouldn't get any hotter?

Ross says: "And the curve for this mathematical transformation should be EXACTLY equivalent to "σ T(2)4" - Eli says it is so !"

As near as i can tell, Eli only ever claimed that the integral for one curve is twice the integral of the other. Such a claim in no way implies that one integral is exactly twice the other at every point.

"What I say is valid mathematics and it proves your assertions about adding up flux and calculating the temperature from the algebraic sums using the SB equation are wrong !"

Why would this invalidate anything? If an object is radiating some amount of power like 267 W/m^2 (ie T=264 K), it will have a specific Planck curve related T=264K. It doesn't matter where it got the power. It might be from 5800 K sunlight. It might be an electric heater. It might be partly from 290 K radiation and partly from 220 K radiation. The outgoing Planck curve does not need to be the sum of the incoming Planck curves at each and every wavelength.

Eli, Your math says that having moving towards having infinite plates would lead to an equilibrium of T=290K for the first plate. What is the difference between a thick object and lots of plates? For example,

What is the difference between 100-molecule thick plate and 100 plates? Is not each column of molecules a plate?

Eli, you violate conservation of energy. Your plate is emitting 200 W/m2 toward the sun at the same time it's receiving 400 W/m2 - so it's really only receiving 200 W/m2.

That's brilliant. By your logic the SB law must always have a 1/2 term. The equipment measuring solar energy must always show HALF of what they actually show, by your logic, because said equipment obviously has another side.

Betty asks: "What is the difference between a thick object and lots of plates?"

Spacing!

With 100 thin plates (for example 0.5 mm thick aluminum spaced 1 mm apart, painted black on each side to make the emissivity close to 1), there is empty space between each, preventing conduction. With a single plate with the same amount of material (50 mm thick with black paint on the outside to match the previous example), conduction will keep both sides of the plate nearly the same temperature.

You are halfway there. Yes, it's receiving 400 and emitting 200 towards the Sun. The other 200 is emitted away from the Sun. This makes energy in = energy out thereby conserving energy as required by law.

If you want 400 to be emitted away from the Sun, construct your plate out of perfectly transparent material. If you want 400 to be emitted towards the Sun, construct your plate out of perfectly reflecting material. If you want 200 to be towards the Sun and 200 to be emitted away from the Sun, construct your plate out of black body material. It's really that simple.

Any instrument measuring flux would of course need to be calibrated such that its own internal physics were not a biasing factor. That's hardly an issue. Or rather, it's an everyday issue with every measuring instrument.

Conduction and convection are not relevant to a discussion of radiation except in that at equilibrium, internally a black body itself is all at the same temp by definition.

or absorbs, conducts, and re-emits. It just refers to emission on the other side after going through the material. You're not arguing that energy doesn't pass through, are you? Does it teleport to the other side?

1. Instruments have many configurations. Your notion here is just plain silly.

2. "As soon as the plate receives 400 from the sun, Eli claims it emits 200 toward the sun. So it only got 200." This is where you clearly show you have no Earthly idea what a black body is.

What anyone who understands the definition would say is that at equilibrium a black body absorbs every bit of incident radiation regardless of direction of incidence and emits an equal amount of radiation equally in all directions. In the case of a plate, only 2 directions are available, of course.

Wrong. Radiation is electromagnetic waves. Once the radiation vibrates the receiving electrons it will jiggle the molecules around. The jiggling of the molecules and electrons doos not automatically lose half its radiation to the very place it got it from. The jiggling would have to slow down for no reason other than your priest, Eli, wants it to.

A black body is a theoretical construct and has no internal physics only a defined external physics. That said, it is an interesting physics which posits that individual molecules generally emit radiation in some preferred directions and not others! They're smart little buggers!

Emissions generally occur in all directions equally. When the emissions come from a vertical plate, half will go to the left and half will go to the right. Even when emissions come from a sphere, down to and including a sphere containing only a single molecule, half will go into one hemisphere and half into the opposite hemisphere.

From this fundamental definitional error follows all of your incredibly wrong-headed thought processes and completely wrong conclusions. Trying to see any sense behind the nonsense you've said, I intuited, as I said above, you seem to be confusing radiation with conduction. Now you've said it. And again, that is the root of your problem.

Conduction requires a physical, material connection. In this case there is no such connection to allow conduction. Nor is there any gaseous or liquid medium to allow convection to proceed. Radiation equations neglect both conduction and convection. Correctly so. Consider the plates to be existing in a vacuum. There simply is neither conduction nor convection across a vacuum gap.

This is also why radiative energy goes both ways. The molecules simply emit radiation in all directions--half to one hemisphere and half to the opposite hemisphere--regardless of which way the energy came in. There is no way a molecule in the present situation can even "know" which way energy came in in the first place. And even if it did it cannot emit directionally or anti-directionally as a consequence of any such "knowledge".

Betty Pound, I am afraid even the kindest of kindest physicist would be unable to answer in a way that would not insult you. After all, it seems that anything that corrects your mistaken views is considered an insult...

Try shining a nice warm heat lamp on anyone's stupid face. Note the molecules don't go anywhere at all. They would only glow a little more brightly in all directions.

As for your ice cubes, where are they? If they are in deep space they will cool to near 0K but will do so a little more slowly than if they had not been near each other. Eli actually kindly answered this waayyy up thread but I guess the kindness didn't penetrate.

Betty, you do actually have a point about the word "equilibrium". Eli should have used the word "steady-state" in the top post.

As stated at Wikipedia "In thermodynamic equilibrium there are no net macroscopic flows of matter or of energy, either within a system or between systems." The net flows of energy that exist in the top post violate this definition. Strictly speaking, the system is NOT in equilibrium.

But only someone being pedantic would make an issue of this. It is clear what was meant -- the system has stopped changing. Furthermore, it is common for people colloquially to say "equilibrium" when the more correct term would be "steady-state".

And blankets block convection, to keep you warm. You seem fixated on convection.

Wait, you don't know that when sun shines on the ground, there is conduction to a few meters? That's sad.

If the ice cubes are surrounded by 0K, then obviously they are not at equilibrium. However, they will cool at the same rate, keeping equilibrium between each other.

Do you even know what radiation is? It's not teleportation. It's an electromagnetic wave. Imagine a rope tied to the object you want to heat. You wave the rope up and down. You claim it's possible for your heated object to dull your hand motion by 50%.

How do you think the 200 W/m^2 gets from the blue plate to the green plate in this case? The green plate is radiating 200 W/m^2 to the right, so it must be supplied with 200 W/m^2 to be at steadystate. But if the two plates are the same temperature (as you imagine), then no heat can flow from blue to green. And there is no other source around.

For heat to get from the blue plate to the green plate, the green plate MUST be cooler than the blue plate.

Once achieving steadystate with the point source (sun), the plates can work on achieving Equilibrium between each other. When they do, Each plate (blue and green) radiates 200 W/m2 left and right. The 200 from blue to green, and 200 from green to blue cancel each other out, just like two ice cubes have 315 W/m2 going both ways.

You are correct to say "The 200 from blue to green, and 200 from green to blue cancel each other out". Thus -- by your own words -- there is no net transfer from blue to green. There is also no net transfer from the sun to green -- no sunlight hits green. There is no net transfer to green at all!

And yet ... there is a 200 net transfer from green to space! The green plate is generating 200 W/m^2 out of thin air -- a clear violation of conservation of energy.

You can't have it both ways! "The moment green loses energy to space, the blue transfers energy to green, to maintain equilibrium.""There is no net transfer to green at all"

EVERY SINGLE MOMENT green is transferring energy to space. Every single movement, blue is transferring energy to green. This is a *steady-state* situation which reaches a constant, fixed temperature difference with a fixed power flow; not an *equilibrium* situation which reaches a constant zero temperature difference with zero energy flow.

Ross says: "And the curve for this mathematical transformation should be EXACTLY equivalent to "σ T(2)4" - Eli says it is so !"

As near as i can tell, Eli only ever claimed that the integral for one curve is twice the integral of the other. Such a claim in no way implies that one integral is exactly twice the other at every point."

You are wrong !

Just how do figure this can work given the explicit relationship betwwen Planck's, the Stefan-Boltzmann and Wein's equations ?

Seriously that is one of the most incompetent statements ever.

This is from the post

σ T(1)^4 = 2 σ T(2)^4 AND σ T(1)^4 is the integral for the Planck curve for T(1).

What is the curve for 2 σ T(2)^4 = σ T(2)^4 + σ T(2)^4 = σ T(1)^4 IF IT IS NOT A PLANCK CURVE ?

If it is NOT a Planck curve then the area under it is NOT σ T(1)^4 and all of the use of the SB equation is invalid!

"I think that all plates will reach "equilibrium" at 244K. Space between plates is not a heater."

Wrong. Yes space is not a heater. But what you don't seem to realize is it's a perfect conductive/convective insulator at the same time which with one plate being in the Sun and the other being in the shade prevents them from being at the same temp. Just like going into the shade on a hot sunny day prevents overheating. You are positing a conductive relationship like if the plates were in contact.

And the problem stipulates that the plates are close by, so the gradient is negligible and therefore nonexistent. Obviously if the plates extended to infinity, it would be odd for them all to be at the first plate's temperature.

Have you worked out the numbers for many plates based on Eli's "physics"? It certainly doesn't progress in "fixed temperature differential" chunks.

I finally may understand your fundamental problem: If the plates are in perfect thermal contact while being illuminated with 400 W/m^2 they exhibit one temp--244K. Now separate them by a slight vacuum space and things change. The green plate cools to 205K as being in the shade it only receives 200W/m^2. The blue plate warms to 262K because of the back radiation of 100W/m^2 from the now colder green plate.

Space is not a "heater" but heat can most definitely can be a redistributed around through it with various ways including half mirrors (which is what a disk black body effectively is). Perhaps if the original article stated the problem in terms of no mirror behind the blue plate, a half mirror (i.e., black body) behind it, and a perfect mirror behind it (leading to a 290K blue plate) you would have understood the situation better?

I did make one error: Solving for the temp of the green plate at equilibrium you get 220K, not 205K--I solved for the starting point not the equilibrium point, stupidly.

Sorry, but the one who is failing to learn is you. Even in space, it matters a lot whether you are surrounded by, but not in conductive contact with, rock at 400C, rock at 200C, rock at 37C, rock at -100C, rock at -200C, or nothing at all. The fact that you don't believe in back radiation or radiation coming from objects colder than other objects means nothing. It's true whether or not you believe it.

It's simply a fact you will radiate away less total heat per unit time when surrounded by 0C rock than when surrounded by near 0K space. You can say it's not true, but you'd be wrong.

On closer reading, you got part but not all of it. 262K is the end point for the blue plate. It doesn't cool from there. It is the point where energy in = energy out.

Following then on through, 220K is the point where energy in = energy out for the green plate.

The point is your criticism is very much on point and will lead you to the right answer if you just follow through the math remembering that energy in always must equal energy out. If there is more going out than coming in, as you consistently have been proposing, you're in the realm of perpetual motion.

1. You are thinking in the physics of conduction not radiation. Wrong PHYSICS. Every object in the universe above 0K radiates to EVERY other object. Radiation does NOT "flow from hot to cold". It is everywhere going to and from all objects in radiative sight of each other. That's just how radiation works. You continually deny this, but it really is a simple, observable fact. The math merely describes the fact.

2. Yes, I added the incoming radiation from being surrounded by but not touching, say, 30C rock in space to your outgoing radiation at 37C (assuming a transparent spacesuit). Correctly so. And I added the lesser incoming radiation from being surrounded by -200C rock and noted the second situation results in you radiating your heat away faster. The backradiation from the 30C vs -200C rock defines the half the radiative situation in the first place. Both are cooler than 37C, but "delay" has nothing to do with it.

3. 220K results from substituting 262K (i.e., T sub 1) back into the original simultaneous equations and solving for T sub 2.

I have just recently come across the topic of GHE and was never aware of how hard it seems to be to grasp even the most trivial points correctly.

I can only see two possibilities:

1) Complete conceptual confusion, which even cannot be solved by a proper physical education.

2) Trying to bend the facts intentively.

I'm not able to tell, which of the two is right in the case of some commenters here, especially Betty Pound.

But it is so simple to grasp the main point, that I must assume possibility 2) to be much more probable.

-----

The discussion NEVER was about, if a colder object is able to heat a warmer one by itself.

The question always and only was, if a constant energy source is able to bring an object A to a higher temperature, as long as another object B is present in its proximity, compared to the different case, where object B is NOT present and object A is completely alone with the energy source. Question: Will the steady-state temperature of A differ from the first case to the second, assuming the energy source is completely constant in any case.

If we assume that radiative interaction between A and B is not prohibited, then in the first case ( energy source heating object A, object B in proximity ) object A MUST have a higher steady-state temperature than in the second case ( energy source heating object A in the nothingness ).

This is not violating any law of physics, on the contrary, it is the only way to satisfy the so far known laws of thermodynamic and radiative energy transfer.

And of course in the first case B is NOT heating A in physical terms ( although you might say so colloquially ).B is slowing the rate of the cooling of A, therefore the constantly incoming energy from the energy source can rise the internal energy of A to a higher level than it would without B in the proximity of A. This results in a higher temperature of A.

Physically spoken, the energy source heats A. A cools radiatively to nothingness. If B is brought in proximity to A, A cools to nothingness AND to B, B heats up in return. Therefore, at steady-state, B will have a certain temperature and thus radiate energy towards A. In return A has then a slower total loss of energy ( since now energy is coming in, that was not coming in before ). This leads finally to a higher temperature, for the constantly energy providing source is unaffected by the exchange between A and B.

So, summing up: The energy source heats A, A cools towards B, B is thus heated by A, as soon as B has a temperature > 0 it will radiate energy. Of this energy some reaches A, thus slowing the radiative energy loss of A, thus the ENERGY SOURCE (!!!) heats up A to a higher steady-state.

So it is the energy source heating everything, not B heating A. Nevertheless A will end up with a higher temperature when B is there compared to the absence of B.

It needs much words to describe this, so a picture is better. But as the comments show, even the most simple picture cannot do its job if the confusion is too big or the intention is to strong.

-----

The only way out for those who seem to deny those effect, is as far as I can see to say that the radiation from B cannot be absorbed by A. Or to break it down to a discussion about semantics, for when it would be right to use the term "heat".

Well, then they have invented new physics and can go have a shot for the Nobel Prize.

Betty: The second law of thermodynamics is merely a summary of the consequences of the detailed, physical, causal mechanisms of the universe. Eli's example and the multiple responses to you by commenters, are explanations of those mechanisms. Correctly understanding and interpreting the summary (the second law) requires understanding the multiple of those detailed, physical, causal, mechanistic explanations, because there are multiple, different, situations needing explanation.

Two of those situations are (1) a system of objects in physical contact, which includes conduction as a mechanism of energy transfer between objects, and (2) a system of objects not in physical contact--perfectly insulated from each other's physical contact by vacuum. Situation 2 includes radiation as the only mechanism of energy transfer. Eli's example is of situation 2, not situation 1.

The behaviors of the systems in both situation 1 and situation 2 legitimately can be summarized by the second law. But the mechanisms within those systems differ.

Your approach is backwards: You are treating the second law like a holy revealed truth, and the detailed, physical, causal, mechanistic explanations like incomplete, rough, approximate explanations. You understand Situation 1's mechanism. You do not understand Situation 2's mechanism, perhaps because you do not recognize the very existence of Situation 2. You are treating Situation 2's mechanism as if it is identical to Situation 1's mechanism. Ironically, you are distorting the meaning of the second law (and/or violating the first law) to accommodate your misunderstanding of Situation 2's mechanism. Those distortions/violations should clue you that your understandings of the detailed physical causal mechanisms are incorrect.

It's not nearly as hard as you are making it. Just accept that there are two situations with two mechanisms. Deal with each of them on their own terms. You will then see that both can be summarized by the second law.

1. You have repeatedly said an ice cube or other colder object does not radiate heat energy to anything warmer. That is neither a hoax nor a misquote nor a misrepresentation of what you've said. But a large ice cube next to you really does radiate more heat your way than, say, a block of dry ice the same size whether you like it or not. Deny it all you want. You are wrong.

2. The very simple reason that the 262K plate cannot warm a nearby plate on the shaded side to more than 220K is that the radiation emitted by said 262K plate on each side is not 400W/m^2 but rather (by a quickie bunch of keypresses on a calculator that I'm open to having corrected) 267W/m^2. This leads to having the green plate emit 133W/m^2 on each side. This means 400 W/m^2 enters the system from the sun and 267 + 133 = 400 W/m^2 exit out the extreme left and extreme right sides of the system. Energy in= energy out. Everyone but you--well and Ross too--is happy.

3. This lower radiation emission rather than your preferred 400W/m^2 emission by each side of the blue plate has been from the beginning and remains till now the essential point you deny. It gets just plain silly for you to continue to deny this as anyone with an honest understanding of even first year physics knows you're wrong. If you were right you would have invented perpetual energy production.

4. The math involved--2 equations with 2 unknowns and taking 4th roots--is quite literally junior high school or early senior high school level depending on your track at the time. It's just not esoteric either.

"Whatever energy the plate received can not be boosted by recycling that energy off of something else to boost itself higher."

NOT receiveD! Constantly recieveS! It makes all the difference.

NOT "boost itself". Nobody ever claimed that anything happens "by itself" here, but by the constant energy supply of a energy source.

If your above citet statement was true, explain one thing: How does clothing warm the human body in winter? The clothing is always colder than the body. The body has its own constant energy supply.If you were right and slowed cooling of a body with a constant energy supply can in no way lead to a higher temperatur of the body, clothing is meaningless and humans can as well be naked during winter.

Alternatively you demand, that the 2nd law has to be applied differently depending on the kind of heat transfer ( conduction, convection, radiation ).

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Eli Rabett

Eli Rabett, a not quite failed professorial techno-bunny who finally handed in the keys and retired from his wanna be research university. The students continue to be naive but great people and the administrators continue to vary day-to-day between homicidal and delusional without Eli's help. Eli notices from recent political developments that this behavior is not limited to administrators. His colleagues retain their curious inability to see the holes that they dig for themselves. Prof. Rabett is thankful that they, or at least some of them occasionally heeded his pointing out the implications of the various enthusiasms that rattle around the department and school. Ms. Rabett is thankful that Prof. Rabett occasionally heeds her pointing out that he is nuts.